How do you integrate #int (5x-1)/(x^2-x-2)# using partial fractions?

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Answer 1 · 2016-05-13T04:25:18

#2 ln (x+1) + 3 ln (x-2) + c#, with #x > 2#

Let #f(x)=(5x-1)/(x^2-x-2)#

#=(5x-1)/((x-2)(x+1))#

#=(A(x-2)+B(x+1))/((x-2)(x+1))#

#=A/(x+1)+B/(x-2)#

Determine matching A and B from #A(x-2)+B(x+1)=5x-1#

Comparing coefficients of x, A+B=5. Comparing constants, #-2A+B=-1#.

Solving, A = 2 and B = 3.

Now, #intf(x) dx#

#= 2 int1/(x+1) dx + 3 int 1/(x-2) dx#

#=2 ln (x+1)+ 3 ln (x-2) + C#

For both the logarithms to be real, #x > 2#